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Solving Rational Equations using the LCD

Solve equations that are fractions on both sides

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Solving Rational Equations using the LCD

A right triangle has leg lengths of 12 and 1x units. Its hypotenuse is 2 units. What is the value of x .

Guidance

In addition to using cross-multiplication to solve a rational equation, we can also use the LCD of all the rational expressions within the equation and eliminate the fraction. To demonstrate, we will walk through a few examples.

Example A

Solve 52+1x=3 .

Solution: The LCD for 2 and x is 2x . Multiply each term by 2x , so that the denominators are eliminated. We will put the 2x over 1, when multiplying it by the fractions, so that it is easier to line up and cross-cancel.

52+1x2x152+2x11x5x+22=3=2x3=6x=x

Checking the answer, we have 52+12=362=3

Example B

Solve 5xx2=7+10x2 .

Solution: Because the denominators are the same, we need to multiply all three terms by x2 .

5xx2(x2)5xx25x2xx=7+10x2=(x2)7+(x2)10x2=7x14+10=4=2

Checking our answer, we have: 5222=7+1022100=7+100 . Because the solution is the vertical asymptote of two of the expressions, x=2 is an extraneous solution. Therefore, there is no solution to this problem.

Example C

Solve 3x+45=6x2 .

Solution: Determine the LCD for 5, x , and x2 . It would be the three numbers multiplied together: 5x(x2) . Multiply each term by the LCD.

3x+455x(x2)13x+5x(x2)14515(x2)+4x(x2)=6x2=5x(x2)16x2=30x

Multiplying each term by the entire LCD cancels out each denominator, so that we have an equation that we have learned how to solve in previous concepts. Distribute the 15 and 4x , combine like terms and solve.

15x30+4x28x4x223x30=30x=0

This polynomial is not factorable. Let’s use the Quadratic Formula to find the solutions.

x=23±(23)244(30)24=23±10098

Approximately, the solutions are 23+100986.85 and 23100981.096 . It is harder to check these solutions. The easiest thing to do is to graph 3x+45 in Y1 and 6x2 in Y2 (using your graphing calculator).

The x -values of the points of intersection (purple points in the graph) are approximately the same as the solutions we found.

Intro Problem Revisit We need to use the Pythagorean Theorem to solve for x .

(12)2+(1x)2=2214+1x2=44x2114+4x211x2x2+44=15x2415=x2x=21515=44x2=16x2

Guided Practice

Solve the following equations.

1. 2xx3=2+3xx29

2. 4x3+5=9x+2

3. 3x2+4x+4+1x+2=2x24

Answers

1. The LCD is x29 . Multiply each term by its factored form to cross-cancel.

2xx3(x3)(x+3)12xx32x(x+3)2x2+6x3xx=2+3xx29=(x3)(x+3)2+(x3)(x+3)13xx29=2(x29)+3x=2x218+3x=18=6

Checking our answer, we have: 2(6)63=2+3(6)(6)29129=2+182743=223

2. The LCD is (x3)(x+2) . Multiply each term by the LCD.

4x3+5(x3)(x+2)4x3+(x3)(x+2)54(x+2)+5(x3)(x+2)4x+8+5x25x305x210x+55(x22x+1)=9x+2=(x3)(x+2)9x+2=9(x3)=9x27=0=0

This polynomial factors to be 5(x1)(x1)=0 , so x=1 is a repeated solution. Checking our answer, we have 413+5=91+22+5=3

3. The LCD is (x+2)(x+2)(x2) .

3x2+4x+4+1x+2(x+2)(x+2)(x2)3(x+2)(x+2)+(x+2)(x+2)(x2)1x+23(x2)+(x2)(x+2)3x6+x24x2+x14=2x24=(x+2)(x+2)(x2)2(x2)(x+2)=2(x+2)=2x+4=0

This quadratic is not factorable, so we need to use the Quadratic Formula to solve for x .

x=1±14(14)2=1±5723.27 and 4.27

Using your graphing calculator, you can check the answer. The x -values of points of intersection of y=3x2+4x+4+1x+2 and y=2x24 are the same as the values above.

Explore More

Determine if the following values for x are solutions for the given equations.

  1. 4x3+2=3x+4, x=1
  2. 2x1x53=x+62x, x=6

What is the LCD for each set of numbers?

  1. 4x, x216
  2. 2x, 6x12, x29
  3. x3, x2x6, x24

Solve the following equations.

  1. 6x+2+1=5x
  2. 53x2x+1=4x
  3. 12x29=8xx32x+3
  4. 6xx21+2x+1=3xx1
  5. 5x34xx+1x+2=1x2+2x
  6. 4xx2+6x+92x+3=3x29
  7. x2x28x+16=xx4+3xx216
  8. 5x2x3+x+1x=6x2+x+122x23x
  9. 3xx2+2x8=x+1x2+4x+2x+1x22x
  10. x+1x2+7x+x+2x23x=xx2+4x21

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